Question

For a particle executing SHM having amplitude 'a' the speed of the particle is one half of its maximum speed when its displacement from the mean position is

• A. $a/2$
• B. a
• C. $a\frac{\sqrt{3}}{2}$
• D. 2a

Answer 1

The correct option is C $a\frac{\sqrt{3}}{2}$

$x=A\sin \omega t\Rightarrow \frac{dx}{dt}=A\omega \cos \omega t\Rightarrow v=A\omega \cos \omega t$
$v=\frac{v_{max}}{2}=\frac{A\omega }{2}$
$\Rightarrow \frac{A\omega }{2}=A\omega \cos \omega t\Rightarrow \cos \omega t=\frac{1}{2}\Rightarrow \sin \omega t=\sqrt{1-{\left(\frac{1}{2}\right)}^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\Rightarrow x=a\sin \omega t=\frac{a\sqrt{3}}{2}$